Derivative of the kinetic energy

AI Thread Summary
Taking the derivative of kinetic energy with respect to velocity results in momentum, as shown by the equation K = 0.5mv^2. The derivative dK/dv simplifies to mv, which aligns with the definition of momentum (p = mv). The discussion highlights the connection between kinetic energy and momentum through calculus concepts. The user confirms that their understanding of derivatives is limited, but they correctly identify momentum as the answer. Overall, the relationship between kinetic energy and momentum is clarified through derivative calculations.
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Derivative of the kinetic energy...

Homework Statement


If you take the derivative of the kinetic energy of a particle with respect to its velocity you get...
A) force
B) momentum
C) acceleration
D) mass
E) potential energy

Homework Equations


K= .5mv^2

The Attempt at a Solution


Im in pre calc right now and i have no idea how to take the derivative of anything, quick google searching led me to believe that its a calculus thing, so it would be really helpfull if someone could explain how to do this..
 
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If you have

y=kx^n

and you want to find the derivative of that with respect to x,denoted as \frac{dy}{dx}

it is simply

\frac{dy}{dx}=knx^{n-1}Where k is a constant
 
I think the answer is B momentum, is that right? momentum has p=mv and kinetic energy has the same variables, so it must be B right?
Also K=mv^(2-1) would be the same as K=mv which is the the same as momentum, is this correct?
 
\frac{dK}{dv}=mv which is momentum so you are correct.
 
You could also exploit the notation and think about it in terms of units.

dK/dv = [\frac{kg * m^2/s^2}{m/s}] = [kg * m/s]
 
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